This is a homework question from my probability class.
Let $X$, $Y$ be two random variables. Suppose $X$ is uniformly distributed on $A$, $Y$ is uniformly distributed on $A$, $A$ $\subseteq$ $\Bbb R^2$. Further assume $X, Y$ are independent.
Prove/Disprove: $A$ is a rectangle with lines parallel to the axes.
Because $X,Y$ are uniformly distributed on $A$ I know that their PDF is equal to:
$f_X(x)=f_Y(y)=1/\lambda(A)$ where $\lambda$ is the Lebesgue measure on the plane.
because $X,Y$ are independent we get:
$f_{(X,Y)}(x,y)= 1/\lambda(A)^2$
Now I got confused... How can I do this?
Thanks!
Since $X$ and $Y$ are independent, $ f_{XY}(x,y)= \begin{cases} f_X(x)f_Y(y), & (x,y)\in A \\ 0, & \text{otherwise} \end{cases}$.
Suppose that $f_{XY}(x,y)>0, \forall (x,y)\in A$. So, for any $x_0$ that $f_X(x_0)>0$, $f_{XY}(x_0,y)>0$ for all $y$ that $f_Y(y)>0$, and $f_{XY}(x_0,y)=0$ for all $y$ that $f_Y(y)=0$, i.e., the support set of non-zero probability for $Y$ remains the same for any arbitrary $x_0$ that $f_X(x_0)>0$. A similar argument applies to $Y$. Hence, $A$ is a rectangle.
The uniformity of distribution is not necessary for $A$ to be a rectangle.